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Please solve RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 1 maths textbook solution.

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Answer:  y =\frac{x^3}{3}+\frac{x^2}{2}+-\log \left | x \right |+C

Hint: You have to integrate by applying integration of xn.

Given: \frac{dy}{dx}=x^2+x-\frac{1}{x},x\neq 0

Solution: \frac{dy}{dx}=x^2+x-\frac{1}{x}

dy=\left (x^2+x-\frac{1}{x} \right )dx

Integrating both sides,

\begin{aligned} &\int d y=\int\left(x^{2}+x-\frac{1}{x}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \frac{1}{x} d x=\log |x|+C\right]\\ &y=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\log |x|+C \mid \end{aligned}

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